Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $p = \dfrac{-5q + 15}{q^2 - 5q + 6} \div \dfrac{-3q + 27}{q - 2} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $p = \dfrac{-5q + 15}{q^2 - 5q + 6} \times \dfrac{q - 2}{-3q + 27} $ First factor the quadratic. $p = \dfrac{-5q + 15}{(q - 2)(q - 3)} \times \dfrac{q - 2}{-3q + 27} $ Then factor out any other terms. $p = \dfrac{-5(q - 3)}{(q - 2)(q - 3)} \times \dfrac{q - 2}{-3(q - 9)} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac{ -5(q - 3) \times (q - 2) } { (q - 2)(q - 3) \times -3(q - 9) } $ $p = \dfrac{ -5(q - 3)(q - 2)}{ -3(q - 2)(q - 3)(q - 9)} $ Notice that $(q - 3)$ and $(q - 2)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac{ -5(q - 3)\cancel{(q - 2)}}{ -3\cancel{(q - 2)}(q - 3)(q - 9)} $ We are dividing by $q - 2$ , so $q - 2 \neq 0$ Therefore, $q \neq 2$ $p = \dfrac{ -5\cancel{(q - 3)}\cancel{(q - 2)}}{ -3\cancel{(q - 2)}\cancel{(q - 3)}(q - 9)} $ We are dividing by $q - 3$ , so $q - 3 \neq 0$ Therefore, $q \neq 3$ $p = \dfrac{-5}{-3(q - 9)} $ $p = \dfrac{5}{3(q - 9)} ; \space q \neq 2 ; \space q \neq 3 $